Phy2020 Test 3 April 11, 2011, Chapters 12-19 (no 15)
All problems worth 3 points total unless stated otherwise – points for parts are given.
g=9.8 m/s2 ; 1 cal = 4.184 J; 1 Hz = 1 cycle/second ; 1000 g = 1 kg ; 100 cm = 1 m;
1 inch=2.54 cm ; 1 Watt = 1 J/s
3
1 liter=1000 cm ; density of water = 1 g/cm3; 1 pound=4.45 N
3 sig figs for all answers except where otherwise stated (-1 point each omission)
Decimal point error is -1/2 credit (except on 6b and 9b)
f=/2 f=1/T v= λf Ideal gas law PV=nRT, where n=number of moles of gas and
R=8.3 J/mole; 1 mole = 6.022 1023 atoms (or molecules)
β (in dB) = 10 log(I/I0), where I0 is the reference sound intensity 10-12 W/m2
Doppler formulas:
for the source moving away from the observer: f’ = f/(1 + vsource/vsound)
for the source toward the observer: f’ = f/(1 - vsource/vsound)
Temperature in Fahrenheit/Centigrade/Kelvin. Water freezes at 32 oF, 0 oC, and 273
K, water boils at 212 oF, 100 oC, and 373 K ; Coulomb’s Law F=kq1q2/r2, k=9 x 109
Nm2/C2, a single electric charge, e, = 1.602 x 10-19 C
1a. (2 points) A boat floating in fresh water (density = 1 g/cm3) displaces 100,000
liters of water. What is the mass of the boat? (Archimedes principle.)
___________________________kg
The weight (=mwaterg) of the water displaced provides the flotation force to
counter the weight (=mboatg) of the boat, so the mass of the boat = mass of the
water displaced, or 100,000 liters * 1,000 cm3/liter * 1 g/cm3 = 108 g = 105
kg=100,000 kg
1b. (1 points) Now the same boat is floating in salt water, density = 1.0270 g/cm3.
How many liters of the salt water does the boat displace? (3 sig figs)
___________________liters
The volume of the salt water displaced is less, since – as we all know – due to
Archimedes principle salt water is more buoyant (a displaced cm3 of salt water
weighs 1.0270 times a displaced cm3 of fresh water and thus has 1.0270 times the
buoyant force.) So, the boat has the same mass, and 100,000/1.0270 liters =
97371 liters of salt water have to be displaced to float the boat.
2. What is the temperature at -40 oC expressed in oF?
-40 oC is 40 oC*1.8 oF/oC = 72 oF below the freezing point of water expressed in
oF units. Water freezes at 32 oF, so 72 OF below that is -40 oF, i. e. the same
number as in oC
3. Underwater, sound moves with a velocity=1480 m/s (faster than in air.) If
successive waves after a particular deep ocean disturbance pass by every 5 seconds
(see picture), what is the wavelength of the waves in m?
_______________________m
velocity of wave = f, where  is the wavelength and f is the frequency.
Since v=1480 m/s and f=1wave every 5 sec=1/5s , then =1480 m/s / (1/5s) = 7400 m
4. How much louder (i. e. by what factor is the sound intensity larger) is 77 dB
sound than a 67 dB sound?
Loudness (sound intensity) is described by the decibel scale,
β (in dB) = 10 log(I/I0), where I0 is the reference sound intensity 10-12 W/m2
Therefore, 77 dB has an I given by 77 = 10 log(I/I0) or 7.7 = log(I/Io) or I/Io=107.7
And 67 dB has an I given by 67 = 10 log(I/I0) or 6.7=log(I/I0) or I/Io=106.7, so the
ratio of the two sound intensities = 107.7/106.7 = 10
5. A source of sound (vsound=343 m/s) is moving away from an observer at sufficient
speed that the emitted sound frequency (1000 Hz) is 30% lower (i. e. 700 Hz) when
detected by the observer. In units of m/s, what is the velocity of the sound emitter?
_______________________________m/s
Use the Doppler formula for the source away from the observer :
f’ = f/(1 + vsource/vsound) and we want f’=0.7 f, so 1+vsource/vsound = 1/0.7=1.429, so
vsource=0.429 vsound =147 m/s
6a. (2 points) A 60 kg woman is walking in high heels. The area of the heel on each
shoe is 1 cm2. If at a particular point of walking she places half of her weight on one
heel, what is the pressure (in N/m2 or Pascal) that the heel exerts on the floor?
_____________________________N/m2
6b. (1 point – no credit for wrong decimal point!) How much is this in a unit you’re
used to, pounds/inch2?
______________________lbs/in2
P=F/A, F=mg, so half of her weight is 30 kg * 9.8 m/s2 = 294 N.
1 cm2=(10-2 m)2 = 10-4 m2; F/A = 294 N/10-4 m2 = 2.94 106 N/m2 or about 430 pounds
per square inch!
R1
A
R2
B
R3
7a. (1 point) If three resistors, R1= 6 Ω, R2= 7 Ω, and R3= 8 Ω are hooked in parallel
between point A and point B (see picture just above), what is the resultant total
resistance of these three resistors? 3 sig figs
1/Rtotal = 1/R1 + 1/R2 + 1/R3 = 0.166667 + 0.1429 + 0.125 = 0.4345 so Rtotal=2.30 Ω
7b. (2 points) If a current of 1 A runs between point A and point B in the parallel
configuration shown in 7a., what is the power dissipated in resistor R2 in units of
Watts? (2 sig figs)
Not all the 1 A of current flows thru each resistor, the current splits up so that the
total flow is 1 A. One way to do this is to calculate the voltage across the 3 resistors
which is the same since they are all in parallel like the lights in your house all see
120 V. So V=IRtotal = 1 A x 2.30 Ω = 2.30 V. So the power loss across resistor R2 is
just V2/R2 = 0.757 W.
Another way to see this, after you have V=2.30 V, is to find the current thru R2
(I=V/R2 = 2.30/7 = 0.3286 A) then, the power dissipated in R2 is just Ithru R22 * R2 =
0.32862 * 7 = 0.756 W (slightly different in the two methods due to rounding)
8. A sealed container has an ideal gas in it. At 20 oC, the pressure in the sealed
container is 104 N/m2. At 100 oC, what is the pressure in the sealed container?
_______________________N/m2
PV=nRT, since the container is sealed, n – the number of moles of the gas – is
constant, as is the volume, V. T=20 oC = 293 K, T=100 oC = 373 K, so the
temperature T increases by a factor of 373/293 = 1.273 so the pressure has to go up
the same amount, so P=1.273 104 N/m2
9a. (2 points) The specific heat of water above the freezing point and below the
boiling point is, to a good approximation, 1 cal/goC. If a water heater for a 50,000
liter swimming pool is to raise the temperature of the water in the pool from 20 oC
to 30 oC in 1 hour, what is the power required?
___________________W
9b. (1 point) If energy costs $0.25/kW-hour (i. e. 25 cents for a power of 1000 W
used for 1 hour), how much money, in $, does it take to heat the pool in a 1 hour
time span using constant power for the whole hour? (no credit for wrong decimal
point)
________________$
50,000 liter = 50,000 l * 1000 cm3/l * 1 g/cm3 = 5 107 g. It takes 1 cal to heat 1 g of
water by 1 oC, and 10 cal to heat 1 g of water by 10 oC from 20 oC to 30 oC. 1 cal =
4.184 J, so heating the swimming pool takes 5 107 g * 10 cal/g * 4.184 J/cal = 20.92
108 J. If this should take place in 1 hour, that’s 1 hr * 60 min/hr * 60 s/min = 3600 s.
So Power = Energy/time = 20.92 108 J/3600 s = 5.81 105 J/s = 5.81 105 W or 581 kW.
This power used for 1 hr gives 581 kW-hr, or $145.28.